(a) Prove that $\sqrt{23}$ is irrational. (b) Prove that for all real numbers $x$ and $y$, where $x^2+y^2 \neq 0$, \[ \frac{(x+y)^2}{x^2+y^2} \leq 2. \] (c) An object with mass $m$ kilograms slides down a smooth inclined plane with velocity $y(t)$, where $t$ is the time in seconds after the object started sliding down the plane. The inclined plane makes an angle $\theta$ with the horizontal, as shown in the diagram. The normal reaction force is $R$ and has magnitude $g$. No other forces act on the object. The vectors $\hat{i}$ and $\hat{j}$ are unit vectors parallel and perpendicular, respectively, to the plane, as shown in the diagram. (i) Show that the resultant force on the object is $\mathbf{F} = - (mg \sin \theta) \hat{j}.$ (ii) Given that the object is initially at rest, find its velocity $y(t)$ in terms of $g$, $\theta$, $t$, and $l.$ (d) Find the cube roots of $2 - 2i$. Give your answer in exponential form. (e) The complex number $2 + i$ is a zero of the polynomial $P(z) = z^4 - 3z^3 + cz^2 + dz - 30$ where $c$ and $d$ are real numbers. (i) Explain why $2 - i$ is also a zero of the polynomial $P(z)$. (ii) Find the remaining zeros of the polynomial $P(z)$.

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(a) Prove that $\sqrt{23}$ is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Question 12

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(a) Prove that $\sqrt{23}$ is irrational. (b) Prove that for all real numbers $x$ and $y$, where $x^2+y^2 \neq 0$, \[ \frac{(x+y)^2}{x^2+y^2} \leq 2. \] ... show full transcript

Worked Solution & Example Answer:(a) Prove that $\sqrt{23}$ is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Step 1

Prove that $\sqrt{23}$ is irrational.

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Answer

Assume for contradiction that (\sqrt{23}) is rational, indicating it can be expressed as (\frac{p}{q}), where (p) and (q) are integers with no common factors, and (q \neq 0). Thus:

\sqrt{23} = \frac{p}{q} \implies 23 = \frac{p^2}{q^2} \implies 23q^2 = p^2.

This implies (p^2) is a multiple of 23, indicating (p) must also be a multiple of 23. Therefore, let (p = 23k) for some integer (k). Substituting this back gives:

23q^2 = (23k)^2 = 529k^2 \implies q^2 = 23k^2.

This further implies (q^2) is also a multiple of 23, hence (q) must be a multiple of 23. Since both (p) and (q) have 23 as a common factor, this contradicts the assumption that they are coprime. Therefore, (\sqrt{23}) is irrational.

Step 2

Prove that for all real numbers $x$ and $y$, where $x^2+y^2 \neq 0$, \(\frac{(x+y)^2}{x^2+y^2} \leq 2.

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Answer

We start by expanding the numerator:

(x+y)^2 = x^2 + 2xy + y^2.

Thus,

\frac{(x+y)^2}{x^2+y^2} = \frac{x^2 + 2xy + y^2}{x^2+y^2} = 1 + \frac{2xy}{x^2+y^2}.

Next, we apply the Cauchy-Schwarz inequality, which states:

(x^2 + y^2)(1 + 1) \geq (x + y)^2 \implies 2(x^2+y^2) \geq (x+y)^2.

Thus, we have:

\frac{(x+y)^2}{x^2+y^2} \leq 2.

Step 3

Show that the resultant force on the object is $\mathbf{F} = - (mg \sin \theta) \hat{j}.$

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Answer

To find the resultant force, we resolve the gravitational force acting on the object into components parallel and perpendicular to the inclined plane.

Considering the weight of the object, (mg), the components can be expressed as:

The component parallel to the incline is (mg \sin \theta).
The component perpendicular to the incline is (mg \cos \theta).

The resultant force (\mathbf{F}) acting on the object is given only by the component parallel to the incline, since the normal force (R) acts oppositely to this force. Thus, we have:

\mathbf{F} = - (mg \sin \theta) \hat{j}.

Step 4

Given that the object is initially at rest, find its velocity $y(t)$ in terms of $g$, $\theta$, $t$, and $l.$

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Answer

Starting with the formula for acceleration:\n

a = \frac{F}{m} = -g \sin \theta.\n$$ Since the object starts at rest, we integrate the acceleration to find velocity:

v(t) = \int a , dt = -g \sin \theta , t + C.\n$$

Given the initial condition (at (t = 0, v = 0)), we find (C = 0). Therefore, the velocity is:

v(t) = -g \sin \theta \, t.\n$$

Step 5

Find the cube roots of $2 - 2i$. Give your answer in exponential form.

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Answer

To find the cube roots, we express (2 - 2i) in polar form:

The modulus is given by:

\sqrt{(2^2 + (-2)^2)} = \sqrt{8} = 2\sqrt{2}.

The argument is: $\tan^{-1}\left(\frac{-2}{2}\right) = \tan^{-1}(-1) = -\frac{\pi}{4}.$

Thus, we can write:
(2 - 2i = 2\sqrt{2} \operatorname{cis}\left(-\frac{\pi}{4}\right)),
where (\operatorname{cis}(\theta) = \cos(\theta) + i\sin(\theta)). \

To find the cube roots:

\text{Roots: } \sqrt[3]{2\sqrt{2}} \operatorname{cis}\left(-\frac{\pi}{4} + \frac{2k\pi}{3}\right), \text{ for } k = 0, 1, 2.

Step 6

Explain why $2 - i$ is also a zero of the polynomial $P(z)$.

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Answer

Since the polynomial (P(z)) has real coefficients, we can apply the complex conjugate root theorem. This theorem states that if a polynomial has real coefficients and one of its roots is a complex number (here, (2 + i)), then the complex conjugate of that root (here, (2 - i)) must also be a root of the polynomial.

Step 7

Find the remaining zeros of the polynomial $P(z)$.

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Answer

Given the polynomial (P(z) = z^4 - 3z^3 + cz^2 + dz - 30) and knowing that (2 + i) and (2 - i) are roots, we can use polynomial long division to factor out ((z - (2 + i))(z - (2 - i)) = (z - 2)^2 + 1).\n We simplify to get:

P(z) = ((z - 2)^2 + 1)(z^2 + az + b)\n$$ Substituting back and solving for \(a\) and \(b\) allows us to find the remaining roots using the quadratic formula.

(a) Prove that \(\sqrt{23}\) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Worked Solution & Example Answer:(a) Prove that \(\sqrt{23}\) is irrational - HSC - SSCE Mathematics Extension 2 - Question 12 - 2023 - Paper 1

Prove that \(\sqrt{23}\) is irrational.

Prove that for all real numbers \(x\) and \(y\), where \(x^2+y^2 \neq 0\), \(\frac{(x+y)^2}{x^2+y^2} \leq 2.

Show that the resultant force on the object is \(\mathbf{F} = - (mg \sin \theta) \hat{j}.\)

Given that the object is initially at rest, find its velocity \(y(t)\) in terms of \(g\), \(\theta\), \(t\), and \(l.\)

Find the cube roots of \(2 - 2i\). Give your answer in exponential form.

Explain why \(2 - i\) is also a zero of the polynomial \(P(z)\).

Find the remaining zeros of the polynomial \(P(z)\).

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